If x=1+sin a/cos a then sin a equala.1-x^2/1+x^2b.x^2-1/x^2+1C.1+X/1-x

1.	If x=1+sin a/cos a then sin a equala.1-x^2/1+x^2b.x^2-1/x^2+1C.1+X/1-xD.1-x/1+x
Answer» $\large\underline{\sf{Solution-}}$ GIVEN that $\rm :\longmapsto\:x = \dfrac{1 + sina}{cosa}$ On squaring both sides, we get $\rm :\longmapsto\: {x}^{2} = \dfrac{ {(1 + sina)}^{2} }{ {cos}^{2} a}$ can be rewritten as $\rm :\longmapsto\:\dfrac{ {x}^{2} }{1} = \dfrac{ {(1 + sina)}^{2} }{ {cos}^{2} a}$ On applying COMPONENDO and Dividendo, we get $\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{ {(1 + sina)}^{2} + {cos}^{2} a }{ {(1 + sina)}^{2} - {cos}^{2} a}$ $\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{1 + {sin}^{2}a + 2sina + {cos}^{2} a }{ 1 + {sin}^{2}a + 2sina - {cos}^{2} a}$ can be re-arranged as $\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{1 + ({sin}^{2}a + {cos}^{2}a) + 2sina }{ (1 - {cos}^{2}a) + {sin}^{2}a + 2sina}$ We know, $\boxed{ \rm{ {sin}^{2}x + {cos}^{2}x = 1}}$ So, USING this $\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{1 + 1+ 2sina }{ {sin}^{2}a + {sin}^{2}a + 2sina}$ $\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{2+ 2sina }{2{sin}^{2}a + 2sina}$ $\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{2(1+ sina)}{2{sina}(sina + 1)}$ $\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{1}{sina}$ $\bf\implies \:sina = \dfrac{ {x}^{2} - 1}{ {x}^{2} + 1}$ Hence, Option (B) is correct Additional INFORMATION :- $\red{\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d} , \: then}$ $\boxed{ \rm{ \frac{a}{c} = \frac{b}{d} \: is \: called \: alternendo}}$ $\boxed{ \rm{ \frac{b}{a} = \frac{d}{c} \: is \: called \: invertendo}}$ $\boxed{ \rm{ \frac{a + b}{b} = \frac{c + d}{d} \: is \: called \: componendo}}$ $\boxed{ \rm{ \frac{a - b}{b} = \frac{c - d}{d} \: is \: called \: dividendo}}$ $\boxed{ \rm{ \frac{a + b}{a - b} = \frac{c + d}{c - d} \: is \: called \: componendo \: and \: dividendo}}$

If x=1+sin a/cos a then sin a equala.1-x^2/1+x^2b.x^2-1/x^2+1C.1+X/1-xD.1-x/1+x

Answer»

$\large\underline{\sf{Solution-}}$

GIVEN that

$\rm :\longmapsto\:x = \dfrac{1 + sina}{cosa}$

On squaring both sides, we get

$\rm :\longmapsto\: {x}^{2} = \dfrac{ {(1 + sina)}^{2} }{ {cos}^{2} a}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{1} = \dfrac{ {(1 + sina)}^{2} }{ {cos}^{2} a}$

On applying COMPONENDO and Dividendo, we get

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{ {(1 + sina)}^{2} + {cos}^{2} a }{ {(1 + sina)}^{2} - {cos}^{2} a}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{1 + {sin}^{2}a + 2sina + {cos}^{2} a }{ 1 + {sin}^{2}a + 2sina - {cos}^{2} a}$

can be re-arranged as

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{1 + ({sin}^{2}a + {cos}^{2}a) + 2sina }{ (1 - {cos}^{2}a) + {sin}^{2}a + 2sina}$

We know,

$\boxed{ \rm{ {sin}^{2}x + {cos}^{2}x = 1}}$

So, USING this

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{1 + 1+ 2sina }{ {sin}^{2}a + {sin}^{2}a + 2sina}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{2+ 2sina }{2{sin}^{2}a + 2sina}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{2(1+ sina)}{2{sina}(sina + 1)}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{1}{sina}$

$\bf\implies \:sina = \dfrac{ {x}^{2} - 1}{ {x}^{2} + 1}$

Hence, Option (B) is correct

Additional INFORMATION :-

$\red{\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d} , \: then}$

$\boxed{ \rm{ \frac{a}{c} = \frac{b}{d} \: is \: called \: alternendo}}$

$\boxed{ \rm{ \frac{b}{a} = \frac{d}{c} \: is \: called \: invertendo}}$

$\boxed{ \rm{ \frac{a + b}{b} = \frac{c + d}{d} \: is \: called \: componendo}}$

$\boxed{ \rm{ \frac{a - b}{b} = \frac{c - d}{d} \: is \: called \: dividendo}}$

$\boxed{ \rm{ \frac{a + b}{a - b} = \frac{c + d}{c - d} \: is \: called \: componendo \: and \: dividendo}}$

If x=1+sin a/cos a then sin a equala.1-x^2/1+x^2b.x^2-1/x^2+1C.1+X/1-xD.1-x/1+x

Additional INFORMATION :-

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